Two unbiased dice are throws together at random. Find the expected value of the total number of points shown up.
(Or) Calculate the expected value of "x", the sum of the scores when two dice are rolled.

Net Answers :

[Expectation: 7 ; Variance: 5.83 ; Standard Deviation: +2.412]

Solution

Let "x" indicate the sum of the points on a die (which is nothing but the number of points on the die)

"X" represents the random variable and P(X = x) represents the probability that the value within the range of the random variable is a specified value of "x"

In the experiment of tossing a dice, there are six possible elementary events, the events of the die showing up either ONE, TWO, THREE, FOUR, FIVE or SIX all of which are mutually exclusive, equally likely and exhaustive.

Therefore the probability of occurance of each elementary event is 1/6
Probabilty that the dice would show up

ONE ⇒ P(ONE)

=

1

6

TWO ⇒ P(TWO)

=

1

6

THREE ⇒ P(THREE)

=

1

6

FOUR ⇒ P(FOUR)

=

1

6

FIVE ⇒ P(FIVE)

=

1

6

SIX ⇒ P(SIX)

=

1

6

Probability for the sum/number of points on the dice to be

1 ⇒ P(X = 1)

=

1

6

2 ⇒ P(X = 2)

=

1

6

3 ⇒ P(X = 3)

=

1

6

4 ⇒ P(X = 4)

=

1

6

5 ⇒ P(X = 5)

=

1

6

6 ⇒ P(X = 6)

=

1

6

The probabilty distribution of "x" would be

x

1

2

3

4

5

6

P(X = x)

1

6

1

6

1

6

1

6

1

6

1

6

Calculations for Mean and Variance

x

P (X = x)

px [x × P (X = x)]

x2

px2[x2 × P (X = x)]

1

1

6

1

6

1

1

6

2

1

6

2

6

4

4

6

3

1

6

3

6

9

9

6

4

1

6

4

6

16

16

6

5

1

6

5

6

25

25

6

6

1

6

6

6

36

36

6

Total

1

21

6

91

6

= 3.5

= + 15.17

Expectation of the sum/number of points on the dice

⇒ Expectation of "x"

⇒ E (x)

=

Σ px

=

3.50

Variance of the sum/number of points on the die

⇒ var (x)

=

E (x2) − (E(x))2

⇒ var (x)

=

Σ px2 − (Σ px)2

=

15.17 − (3.5)2

=

15.17 − 12.25

=

2.92

Let x1, x2 represent the sum of the points on the first die and the second dice respectivley

Expected sum/number of the points

On the First Dice
⇒ E (x1) = 3.5

On the Second Dice
⇒ E (x2) = 3.5

Each trial (throwing of the dice) is identical and therefore the expected sum/number of points on the dice in each trial would be the same

Therefore, expected sum of the points on the two dice

= E (x1) + E (x2)
= 3.5 + 3.5
= 2 × 3.5
= 7

Variance of the sum/number of points

On the First Dice
⇒ var (x1) = 2.92

On the Second Dice
⇒ var (x2) = 2.92

Each trial (throwing of the dice) is identical and therefore the variance of the sum/number of points on the dice in each trial would be the same